How do you find the indefinite integral of #int x(5^(-x^2))#?

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Answer 1 · 2016-12-12T11:51:31

The answer is #=-5^(-x^2)/(2ln5)+C#

We do the integral by substitution

Let #u=x^2#, then, #du=2xdx#

#intx(5^(-x^2))dx=1/2int(5^(-u))du#

Let #v=5^(-u)#

#lnv=ln5^(-u)=-u ln5#

#v=e^(-u ln5)#

Therefore,

#intx(5^(-x^2))dx=1/2int(5^(-u))du=1/2inte^(-u ln5)du#

#=-e^(-u ln5)/(2ln5) =-5^(-u)/(2ln5)=-5^(-x^2)/(2ln5)+C#